Does the following function series converge?

Question

Let $$ f_n(x)=\frac{\frac{1}{(n-1)!}\sum_{k=0}^{\lfloor \alpha n-x\rfloor}C_{n-1}^{k}~(-1)^k(\alpha n-x-k)^{n-1}}{\frac{1}{n!}\sum_{k=0}^{\lfloor \alpha n\rfloor}C_{n}^{k}(-1)^k(\alpha n-k)^{n}}, $$ where

• $x\in[0,1]$,
• $C_{n}^{k}$ is the binominal coefficient,
• and $\alpha$ is a constant such that $0 \le \alpha \le 1$.

Based on my intuition and numerical results, I guess the above series converges pointwise to a truncated exponential function $g(x)=A\exp(-\lambda x)$, where $A$ and $\lambda$ are parameters to be determined.

Question: How to prove or disprove this conjecture?

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This question originates from my studies on the the marginal distribution of a uniform distribution defined over an $n$-dimensional simplex truncated by a unit cube, which can be defined as $$ \mathscr{T}_n(t)=\bigg\{\vec{\mathbf{x}}:\sum_{i=1}^n x_i \le t, 0 \le x_i \le 1\bigg\}. $$ Considering a uniform distribution over the domain $\mathscr{T}_n(\alpha n)$, I obtained the density function of the marginal distribution in any dimension as $$ p(x)=f_n(x)=\frac{\text{vol}\left(\mathscr{T}_{n-1}\left(\alpha n-x\right)\right)}{\text{vol}\left(\mathscr{T}_{n}\left(\alpha n\right)\right)},$$ where $x\in[0,1]$. It is known that the marginal distribution of the joint random vector uniformly distributed over a simplex with a finite and nonzero mean value will converge to an exponential distribution. For this reason, I guess the considered series converges to the density function of a truncated exponential distribution.

Answer 1

$\newcommand{\si}{\sigma}$ By the Irwin--Hall formula, your first displayed ratio is \begin{equation} f_n(x)=\frac{P(S_{n-1}\le an-x)}{P(S_n\le an-x)}=\frac{P(S_{n-1}\le a(n-1)-(x-a))}{P(S_n\le an-x)}, \end{equation} where $a:=\alpha\in[0,1]$, $x\ge0$, $S_n:=X_1+\dots+X_n$, and $X_1,\dots,X_n$ are iid random variables each uniformly distributed on $[0,1]$.

If $a=0$ then $P(S_n\le an-x)=0$ for $x\ge0$, so that $f_n(x)$ is undefined. If $a>1/2$ then, by the law of large numbers, $P(S_n\le an-y)\to1$ (as $n\to\infty$) for any fixed real $y$, so that $f_n(x)\to\frac11=1$. If $a=1/2$ then, by the central limit theorem, $P(S_n\le an-y)\to1/2$ for any fixed real $y$, so that $f_n(x)\to\frac{1/2}{1/2}=1$.

It remains to consider the nontrivial case when $a\in(0,1/2)$. Since $X_i$ equals $1-X_i$ in distribution, we have \begin{equation} f_n(x)=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)}, \end{equation} where \begin{equation} b:=1-a\in(1/2,1). \end{equation} By Theorem 1 by Petrov, \begin{equation} P(S_n\ge tn)\sim\frac{e^{nL_t(h_t)}}{h_t\si(h_t)\sqrt{2\pi n}} \tag{*} \end{equation} uniformly in $t$ in any closed subinterval of the interval $(1/2,1)$, where \begin{equation} L_t(h):=-ht+\ln R(h),\quad R(h):=Ee^{hX_1}=\frac{e^h-1}h,\quad\si(h):=m'(h),\quad m(h):=R'(h)/R(h) \end{equation} for real $h>0$, and $h_t\in(0,\infty)$ is the only root of the equation \begin{equation} m(h_t)=t. \end{equation}

The functions $m$ and $\si$ (on $(0,\infty)$) are smooth, and $\si>0$. So, $m$ is a smooth increasing function, and hence the function $(1/2,1)\ni t\mapsto h_t$ is smooth. So, if $t\to t_0\in(1/2,1)$, then \begin{equation} h_t\si(h_t)\sim h_{t_0}\si(h_{t_0}) \end{equation} and \begin{equation} \frac d{dt}L_t(h_t)=\frac{\partial L_t(h)}{\partial h}\Big|_{h=h_t}\;\frac{dh_t}{dt}-h_t =(-t+m(h_t))\;\frac{dh_t}{dt}-h_t=-h_t\sim-h_{t_0}, \end{equation} whence, by (*), \begin{equation} \frac{P(S_n\ge tn)}{P(S_n\ge t_0n)}\sim \exp[-nh_{t_0}(t-t_0)(1+o(1))]. \end{equation} Using this with $t_0=b$ and $t=b+x/n$, we get \begin{equation} \frac{P(S_n\ge bn+x)}{P(S_n\ge bn)}\sim e^{-h_b x} \end{equation} for each real $x$. Hence, \begin{align} P(S_n\ge bn)&=\int_0^1 P(S_{n-1}\ge bn-z)\,dz \\ &=\int_0^1 P(S_{n-1}\ge b(n-1)+b-z)\,dz \\ &\sim P(S_{n-1}\ge b(n-1))\int_0^1 e^{-h_b(b-z)}\,dz \\ &=P(S_{n-1}\ge b(n-1))e^{-h_b b}R(h_b). \end{align}

We conclude that \begin{align} f_n(x)&=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)} \\ &=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_{n-1}\ge b(n-1))} \frac{P(S_{n-1}\ge b(n-1))}{P(S_n\ge bn)} \frac{P(S_n\ge bn)}{P(S_n\ge bn+x)} \\ &\sim e^{-h_b(x-a)}\frac{e^{h_b b}}{R(h_b)}\,e^{h_b x} =\frac{e^{h_b}}{R(h_b)} \end{align} for each real $x$.

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For an illustration, here are the graphs $\{(x,f_n(x)/\frac{e^{h_b}}{R(h_b)})\colon|x|<5\}$ with $a=0.25$ for $n=100$ (left) and $n=500$ (right):